Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{-2k + 10}{-8k - 24} \div \dfrac{k^2 - 7k + 10}{-2k - 6} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{-2k + 10}{-8k - 24} \times \dfrac{-2k - 6}{k^2 - 7k + 10} $ First factor the quadratic. $x = \dfrac{-2k + 10}{-8k - 24} \times \dfrac{-2k - 6}{(k - 5)(k - 2)} $ Then factor out any other terms. $x = \dfrac{-2(k - 5)}{-8(k + 3)} \times \dfrac{-2(k + 3)}{(k - 5)(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -2(k - 5) \times -2(k + 3) } { -8(k + 3) \times (k - 5)(k - 2) } $ $x = \dfrac{ 4(k - 5)(k + 3)}{ -8(k + 3)(k - 5)(k - 2)} $ Notice that $(k + 3)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 4\cancel{(k - 5)}(k + 3)}{ -8(k + 3)\cancel{(k - 5)}(k - 2)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $x = \dfrac{ 4\cancel{(k - 5)}\cancel{(k + 3)}}{ -8\cancel{(k + 3)}\cancel{(k - 5)}(k - 2)} $ We are dividing by $k + 3$ , so $k + 3 \neq 0$ Therefore, $k \neq -3$ $x = \dfrac{4}{-8(k - 2)} $ $x = \dfrac{-1}{2(k - 2)} ; \space k \neq 5 ; \space k \neq -3 $